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In order to write down the Killing equation, if by definition a vector field $X$ is said to be Killing $\iff$ $L_X g=0$, then I can rewrite this condition as: $$L_X g=X g(U, V)-g(L_XU, V)-g(U,L_XV)=g(\nabla_XU, V)+g(U, \nabla_XV)-g(L_XU,V)-g(U,L_XV)=g(\nabla_XU-L_XU, V)+g(U, \nabla_X V-L_XV)=g(\nabla_U X, V)+g(U, \nabla_V X)=0$$ Now my question is: I have read always the Killing equation as $\nabla_U X+\nabla_V X=0$, but what is the passage from "$g(\nabla_U X, V)+g(U, \nabla_V X)=0$" to "$\nabla_U X+\nabla_V X=0$"?
And how it can be translated in coordinates as $X_{\alpha;\beta}+X_{\beta;\alpha}=0$ (this is another expression I have found out)?

$\textbf{EDIT with my work:}$ I choose to write all in coordinates, so $X=X^{\alpha}\partial_\alpha$ and then $$L_X g_{\sigma \beta}=X^{\alpha}\partial_\alpha g_{\sigma \beta}-g( \partial_\sigma, [X^{\alpha}\partial_\alpha, \partial_\beta])=X^{\alpha}\color{red}{\nabla_{\alpha} g_{\sigma \beta}}-g(X^\alpha[\partial_\alpha, \partial_\sigma]-(\partial_\sigma X^\alpha)\partial_\alpha,\partial_\beta)-g(\partial_\sigma, x^\alpha [\partial_\alpha, \partial_\beta]-(\partial_\beta X^\alpha)\partial_\alpha)=\color{lightgreen} 0+g_{\alpha \beta}\partial_\sigma X^\alpha+g_{\sigma\alpha}\partial_\beta X^\alpha=\color{pink}{\partial_\sigma g_{\alpha \beta}X^\alpha}+\color{violet}{\partial_\beta g_{\sigma\alpha} X^\alpha}=\nabla_\sigma X_{\beta}+\nabla_{\beta}X_{\sigma}=X_{\beta;\sigma}+X_{\sigma;\beta}$$

So my questions become:

$\textbf{1)}$ The passages are right?

$\textbf{2)}$What I have written in red is right? I have thought that $g_{\sigma\beta}$ can be seen as a scalar function for which so the covariant derivative coincides with the partial one.

$\textbf{3)}$In the pink and violet terms I have put into the partial derivatives the terms $g_{\alpha\beta}$ and $g_{\sigma\alpha}$, since the partial derivative is linear, right?

$\textbf{4)}$ The expression obtained is $X_{\beta;\sigma}+X_{\sigma;\beta}$ and not $X_{\beta;\alpha}+X_{\alpha;\beta}$ as in my book...where I am failing?

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    $\begingroup$ Instead of $\nabla_U X+\nabla_V X=0$ I expect you meant $\nabla_\mu X_\nu+\nabla_\nu X_\mu=0$ $\endgroup$
    – octonion
    Commented Mar 14, 2021 at 1:22
  • $\begingroup$ Oh yes sorry! I am really confused...can you give me all passages in coordinates form $L_Xg=0$ to $X_{\nu;\mu}+X_{\mu;\nu}$, please? Thanks! $\endgroup$
    – Nik
    Commented Mar 14, 2021 at 7:42
  • $\begingroup$ I have edited the question with my works...I have not received answers so maybe my question is too trivial, now that I have added my works can you help me please? $\endgroup$
    – Nik
    Commented Mar 14, 2021 at 8:50

1 Answer 1

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You got the result but some steps are not correct. Here is how I would have done it:

The components of the Lie derivative of a 2-covariant tensor are

$$(L_X T)_{\mu\nu}=X^\lambda \partial_\lambda T_{\mu\nu}+T_{\lambda\nu}\partial_\mu X^\lambda+T_{\mu\lambda}\partial_\nu X^\lambda.\tag{1}$$

It can be proved that if there is no torsion, i.e. the connection is symmetric, all partial derivatives in $(1)$ can be replaced by covariant derivatives.

$$\Gamma^\mu_{\nu\lambda}=\Gamma^\mu_{\lambda\nu}\Leftrightarrow(L_X T)_{\mu\nu}=X^\lambda \nabla_\lambda T_{\mu\nu}+T_{\lambda\nu}\nabla_\mu X^\lambda+T_{\mu\lambda}\nabla_\nu X^\lambda \tag{2}$$

Besides, the covariant derivative of the metric tensor is $0$, so $$(L_X g)_{\mu\nu}=X^\lambda \nabla_\lambda g_{\mu\nu}+g_{\lambda\nu}\nabla_\mu X^\lambda+g_{\mu\lambda}\nabla_\nu X^\lambda=\nabla_\mu(g_{\lambda\nu}X^\lambda)+\nabla_\nu(g_{\mu\lambda}X^\lambda)=\\=\nabla_\mu X_\nu+\nabla_\nu X_\mu,$$

then $$L_X g=0\Leftrightarrow\nabla_\mu X_\nu+\nabla_\nu X_\mu=0.$$

Regarding your last questions:

  1. $g_{\mu\nu}$ can't be seen as a scalar function.

  2. The partial derivative is linear, right.

  3. You started with $\sigma$ and $\beta$ as free indices, I chose $\mu$ and $\nu$, the book chose $\alpha$ and $\beta$...

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  • $\begingroup$ Thanks a lot! 1)So the fact that I can consider $\partial_{\alpha}g_{\sigma\beta}=\nabla_{\alpha}g_{\sigma\beta}$ is due by the fact that we are considering as $\nabla$ the levi civita connection and so there is no torsion? 2) Another question: in order to have the expression of the book it is sufficient doing all my passages and only substituting, in the expression $X_{\beta;\sigma}+X_{\sigma;\beta}$, $\sigma=\alpha$? $\endgroup$
    – Nik
    Commented Mar 14, 2021 at 9:38
  • $\begingroup$ When the Levi Civita connection is considered, $\nabla_\alpha g_{\sigma \beta}$ is always $0$, so $\partial_\alpha g_{\sigma\beta}\neq\nabla_\alpha g_{\sigma \beta}$. What happens is that all partial derivatives in (1) can be replaced by covariant derivatives (this is true for the Lie derivative of anything as long as there is no torsion). And yes, it is sufficient to replace $\sigma$ with $\alpha$, as $\sigma$ is a free index. $\endgroup$
    – AFG
    Commented Mar 14, 2021 at 9:44
  • $\begingroup$ Sorry for the stupid questions that I am doing, but if $\partial_{\alpha}g_{\sigma\beta}\neq \nabla_{\alpha}g_{\sigma\beta}$, how it is possible to rewrite $L_Xg=\partial_{\alpha}g_{\sigma\beta}+...$ as $L_Xg=\nabla_{\alpha}g_{\sigma\beta}+...$? It is sufficient to say that the partial derivative can be replaced by the covarinat one? Thanks for the patience! $\endgroup$
    – Nik
    Commented Mar 14, 2021 at 9:55
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    $\begingroup$ I have edited my answer to make it clearer adding (2). You're welcome! $\endgroup$
    – AFG
    Commented Mar 14, 2021 at 10:11
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    $\begingroup$ You can also work exactly from (1) and derive the killing equation explicitly, without needing to use other known results. You just need to rewrite the partial derivatives in terms of covariant derivatives and connection terms. $\endgroup$
    – Eletie
    Commented Mar 14, 2021 at 10:13

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